a = 6s^2
da/dt = 12s ds/dt
da/dt = 12*5*2 = 120 in^2/min
At a certain instant, each edge of a cube is 5 inches long and the volume is increasing at the rate of 2 cubic inches per minute. How fast is the surface area of the cube increasing?
2 answers
V = s^3
dV/dt = 3v^2 ds/dt
when s = 5, dV/dt = 2
2 = 3(25) ds/dt
ds/dt = 2/75
A = 6s^2
dA/dt = 12s ds/st
= 12(5) (2/75) = 8/5 inches^2 / min
dV/dt = 3v^2 ds/dt
when s = 5, dV/dt = 2
2 = 3(25) ds/dt
ds/dt = 2/75
A = 6s^2
dA/dt = 12s ds/st
= 12(5) (2/75) = 8/5 inches^2 / min
2 answers